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Question

If one zero of the polynomial$$ \left( { a }^{ 2 }+9 \right) { x }^{ 2 }+13x+6a$$ is reciprocal of the other, find the value of a


Solution

$$\Rightarrow$$  The given quadratic polynomial is $$(a^2+9)x^2+13x+6a$$.
$$\Rightarrow$$  Let one zero of the quadratic polynomial be $$\alpha$$.
$$\therefore$$  The other zero of the quadratic polynomial is $$\dfrac{1}{\alpha}.$$
$$\Rightarrow$$  $$\alpha\times \dfrac{1}{\alpha}=\dfrac{6a}{a^2+9}$$
$$\Rightarrow$$  $$1=\dfrac{6a}{a^2+9}$$
$$\Rightarrow$$  $$a^2+9=6a$$
$$\Rightarrow$$  $$a^2-6a+9=0$$
$$\Rightarrow$$  $$(a-3)^2=0$$
$$\Rightarrow$$  $$a-3=0$$
$$\Rightarrow$$  $$a=3$$
$$\therefore$$  The value of $$a$$ is $$3$$.  


Mathematics

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